5 Variable K-Map.

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Presentation transcript:

5 Variable K-Map

How many k-map is needed? If you have 5 variables, you’ll need 2 k-map… Let’s say the variables are A, B, C, D and E. 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 E = 0

How many k-map is needed? 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10

Try this out… Simplify the Boolean function F(A,B,C,D,E) = (0,1,4,5,16,17,21,25,29) Soln: F(A,B,C,D,E) = A’B’D’+AD’E+B’C’D’

1st step – convert the minterm into Boolean equation F(A,B,C,D,E) = (0,1,4,5,16,17,21,25,29) How to convert … ?? 0 = 00000  1 = 00001 

1st step – convert the minterm into Boolean equation F(A,B,C,D,E) = (0,1,4,5,16,17,21,25,29)

2nd – prepare 2 k-map E = 0 E = 1 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0 E = 1

3rd- plug in the Boolean term/minterm into k-map 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1

4th- look for similar grouping that can be done in both k-map 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1 That will give us

4th- look for similar grouping that can be done in both k-map 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1 That will give us

The combination will look like this… 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1 That will give us

5th- look for the remaining 1’s that has not been included yet 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1

6th- group the remaining 1’s 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1 That new grouping will give us Note that this time, E need to be included in the term, since the grouping is in the E = 1 k-map only.

6th- group the remaining 1’s 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1 Full expression will be,